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frozen [14]
3 years ago
14

Can someone please help me​

Physics
1 answer:
mario62 [17]3 years ago
8 0

Answer:

GE = ME - \frac{1}{2} m\,v^2,  which agrees with option C in your list.

Explanation:

The definition of Mechanical Energy (ME) of a system is the addition of the gravitational potential energy (GE) plus the kinetic energy (KE) of the system:

ME = GE + KE

Given that the KE is: \frac{1}{2} m\,v^2,

solving for GE in the formula above gives:

GE = ME - KE = ME - \frac{1}{2} m\,v^2,  which agrees with option C

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A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)
AVprozaik [17]
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time 
The vo=0 and vf = 25 m/s and t=7 sec. This gives... 
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N. 
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again). 
5 0
3 years ago
Read 2 more answers
A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
4 years ago
he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
S_A_V [24]

Answer:

207.4 N

Explanation:

The torque \tau  on a body is

\tau = r* F  where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

\tau = rF\sin \theta

Where \theta is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation \tau = rF\sin \theta

\tau = LF\sin \theta

Where L is the length of the wrench.

Making F the subject

F = \frac{\tau }{{L\sin \theta }}

Force required to pull the wrench is given as,

F = \frac{\tau }{{L\sin \theta }}

Substitute 47{\rm{ N}} \cdot {\rm{m}}  for \tau, 25 cm for L, and 115o for \theta  

\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}  

6 0
3 years ago
A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes
mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

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d_1 = v* t

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now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

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d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

3 0
3 years ago
In an electronic transition an atom can not emit what?
iren [92.7K]
█ Question <span>█

</span><span>In an electronic transition, an atom cannot emit what?

</span>█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light. 

<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>
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